Does an electromagnetic gauge transform induce a $U(1)$ transform on the field?

Question

For the free complex scalar Lagrangian, $$\mathscr{L}=\partial_\mu \phi\partial^\mu\phi^{\dagger}-m^2 \phi \phi^{\dagger} $$ if we want it to be invariant under a transformation of the form $\phi\rightarrow e^{iq\alpha(x)}\phi$; $\phi^\dagger\rightarrow e^{-iq\alpha(x)}\phi^\dagger$ we can introduce an operator $$D_\mu=\partial_\mu-iq\partial_\mu(\alpha(x))$$ This would make the Lagrangian invariant under the mentioned $U(1)$ transformation. If we introduce a free electromagnetic field, then $$\mathscr{L}_{EM}=-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}$$ is itself invariant under the electromagnetic gauge transformation $A_\mu\rightarrow A_\mu+\partial_\mu \Lambda$. Then why do the definitions of the covariant derivative involve the electromagnetic gauge factor as $$\mathcal{D}_\mu=\partial_{\mu} +iqA_\mu$$ I mean does the electromagnetic gauge transformation induce the $U(1)$ complex scalar field gauge transformation in some manner? Are $\alpha(x)$ and $\Lambda$ somehow connected or can we apply them as two separate transformations for free fields? Does the inclusion of interaction terms enforce any connection between the two seemingly different gauge transformations?

Answer 1

As pointed out by @Nihbar Karve in the comments, the formula $D_\mu = \partial_\mu - iq\partial_\mu \alpha$ makes no sense, since $\alpha$ only appears when you are doing the transformation.

Starting from the initial lagrangian, we see that the kinetic term is not gauged invariant. We want to introduce a new derivative $D_\mu$ which could be gauge invariant. Since $\partial_\mu$ and $D_\mu$ are derivatives (satisfy the Leibniz rule), we can derive that they only differ by a $4$-vector $A_\mu$, ie : $$D_\mu = \partial_\mu + iqA_\mu$$

(where the $i\alpha$ is arbitrary, but does change much).

For $D_\mu$ to be gauge-invariant, we need it to transform as : $$D_\mu \longrightarrow D_\mu - iq\partial_\mu \alpha$$ This is the case if, and only if, $A_\mu$ transforms as : $$A_\mu\longrightarrow A_\mu - \partial_\mu\alpha$$

What lagrangian we then choose for the gauge field $A_\mu$ is, at that point, arbitrary, though $\mathcal L_{EM}$ is the simplest non-trivial one.

Answer 2

First we must clear up what do we mean by "induce". The fields $A_\mu$ and $\phi$ are independent fields. In that sense there is no way in which transforming $A_\mu$ automatically transforms $\phi$ as a by-product. If you ask about "inducing the transformation" in this sense the answer would be no. But if we understand "induce" to mean that given the transformation of $A_\mu$ or $\phi$ we can find out how the other must transform to keep the action invariant then the answer would be yes and there are two complementary viewpoints.

Usually one starts with the free scalar action, observes that it is invariant under $\phi\mapsto e^{iq\alpha}\phi$ when $\alpha$ is constant and that when $\partial_\mu\alpha\neq0$ it is no longer invariant. What you observe is that the whole point is the derivative. If you can create a new derivative operator $D_\mu$ such that $D_\mu \phi\mapsto e^{iq\alpha} D_\mu \phi$ even when $\alpha$ is not constant, the same arguments for the constant $\alpha$ case would lead you to the invariance under the local symmetry.

You then write such $D_\mu$ in terms of a new field $A_\mu$ and looking at the transformation of the action guess how $A_\mu$ must transform to give you the desired property of $D_\mu$. Later you introduce a kinetic term for the $A_\mu$ field, which obviously cannot break the symmetry you worked all this way to ensure. Given the transformation of $A_\mu$ the most simple kinetic term is the Maxwell term.

That is the rationale beyond this usual story of "we want to make the symmetry local". If anything you could say that the transformation of $\phi$ "induced" that of $A_\mu$ where by "induced" we should mean that it tells how $A_\mu$ should transform if we want the action to be invariant.

A complementary viewpoint is proposed by Weinberg in "The Quantum Theory of Fields". Usually we want to encode creation and annihilation operators of relativistic particles on relativistic fields in order to easily construct interactions respecting Lorentz symmetry. Still, if you try to encode the creation and annihilation operators of photons into a vector field $A_\mu$ we find that it doesn't work. The field $A_\mu$ does not transform correctly as a vector field. Rather it transforms as a vector field up to a gauge transformation.

The solution is to demand that the action be invariant under $A_\mu \mapsto A_\mu + \partial_\mu \alpha$ so that for all purposes $A_\mu$ transforms as a vector. The next question would be about the interactions. Weinberg argues that the interactions must be of the form $A_\mu J^\mu$ where $\partial_\mu J^\mu=0$. Then he recalls how to get such $J^\mu$, and that would be by considering matter fields whose action exhibits the ${\rm U}(1)$ symmetry. In this scenario you could say that the transformation of $A_\mu$, which appeared through the study of embedding photons into vector fields, induce the ${\rm U}(1)$ transformation of the matter fields.

Answer 3

From the answer by @SolubleFish and discussion with @NiharKarve, I have come to the following conclusion about my question. Firstly, the definition of covariant derivative I suggested doesn't make sense as in the definition, we use an arbitrary function $\alpha(x)$ which I adopt from the gauge transform on the complex scalar field and hence can't be used in the definition of the covariant derivative, as pointed by @SolubleFish.

Finally, as per the given definition, $$\mathcal{D}_\mu=\partial_{\mu} +iqA_\mu$$ there may arise three cases:

1. We introduce a gauge transformation $\phi\rightarrow e^{iq\alpha(x)}\phi$; $\phi^\dagger\rightarrow e^{-iq\alpha(x)}\phi^\dagger$, the derivative term acts as $$\mathcal{D}_\mu(e^{iq\alpha(x)}\phi)=e^{iq\alpha(x)}[\partial_\mu \phi+iq\phi(\partial_\mu \alpha(x)+A_\mu)]$$ This is equal to $e^{iq\alpha(x)}\mathcal{D_\mu}\phi$ under the premise that we absorb the $\partial_\mu\alpha(x)$ factor as a gauge transformation of $A_\mu$. This way in order to keep the to total Lagrangian ($\mathscr{L+L}_{EM}$) invariant.

2. Under a gauge transform of the vector potential $A_\mu\rightarrow A_\mu+\partial_\mu \Lambda$, we know that the electromagnetic term would be invariant. The covariant derivative term becomes $$\mathcal{D_\mu}\phi=\partial_\mu \phi+iqA_\mu \phi+iq\partial_\mu\Lambda \phi$$ Multiplying both sides by $e^{iq\Lambda(x)}$, we get $$e^{iq\Lambda(x)}\mathcal{D_\mu}\phi=e^{iq\Lambda(x)}[\partial_\mu \phi+iqA_\mu \phi+iq\partial_\mu\Lambda \phi]=\mathcal{D_\mu(e^{iq\Lambda(x)}\phi)}$$ Hence to maintain invariance of the complete lagrangian, gauge transformation of the vector potential forces us to make a change to the complex scalar field via a $U(1)$ gauge factor.

3. When we do both of the gauge transformations, $\phi\rightarrow e^{iq\alpha(x)}\phi$; $\phi^\dagger\rightarrow e^{-iq\alpha(x)}\phi^\dagger$; $A_\mu\rightarrow A_\mu+\partial_\mu \Lambda$, to maintain invariance of the complete lagrangian, we have to absorb the change in complex scalar field into the vector potential and vice versa, resulting into the total gauge transformation $\phi\rightarrow e^{iq(\alpha(x)+\Lambda(x))}\phi$; $\phi^\dagger\rightarrow e^{-iq(\alpha(x)+\Lambda(x))}\phi^\dagger$; $A_\mu\rightarrow A_\mu+\partial_\mu (\Lambda+\alpha)$.

Answer 4

Yes, $\alpha(x)$ and $\Lambda$ are connected because the electrognetic field only makes sense if it acts on charges.

Answer 5

The Lagrangian of a scalar QED is given by \begin{align} \mathcal{L}&=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+(D_{\mu}\phi)^{\ast}(D^{\mu}\phi)-m^{2}|\phi|^{2} \\ &=\frac{1}{2}(|\mathbf{E}|^{2}-|\mathbf{B}|^{2})+(D_{\mu}\phi)^{\ast}(D^{\mu}\phi)-m^{2}|\phi|^{2} \tag{1} \end{align}

where the convention $D_{\mu}\phi=\partial_{\mu}\phi-iA_{\mu}\phi$ is used. The equations of motion are: $$D_{\mu}D^{\mu}\phi+m^{2}\phi=0\quad\mathrm{and}\quad\partial_{\mu}F^{\mu\nu}=j^{\nu} \tag{2},$$

where $J^{\mu}=i\left[(D^{\mu}\phi)^{\ast}\phi-\phi^{\ast}(D^{\mu}\phi)\right].$

Global $U(1)$-Symmetry:

Under the transformation $$\phi(x)\rightarrow e^{i\Lambda}\phi(x),\quad\phi(x)^{\ast}\rightarrow\phi(x)^{\ast}e^{-i\Lambda}$$

the action (1) is invariant. The Noether current associated with this global $U(1)$-symmetry is given by $$J^{\mu}=i\left[(D^{\mu}\phi)^{\ast}\phi-\phi^{\ast}(D^{\mu}\phi)\right], \tag{3}$$

which is conserved on-shell.

Local $U(1)$-Gauge Invariance:

The Lagrangian density (1) is invariant under $$\phi(x)\rightarrow e^{i\Lambda(x)}\phi(x),\quad\mathrm{or}\quad\delta\phi(x)=i\phi(x)\delta\Lambda(x) $$ $$\phi(x)^{\ast}\rightarrow\phi(x)^{\ast}e^{-i\Lambda(x)}\quad\mathrm{or}\quad\delta\phi(x)^{\ast}=-i\phi(x)^{\ast}\delta\Lambda(x)$$ $$A^{\mu}(x)\rightarrow A^{\mu}(x)+\partial^{\mu}\Lambda(x)\quad\mathrm{or}\quad\delta A^{\mu}(x)=\partial^{\mu}\delta\Lambda(x)$$

It is clear that the Noether current $J^{\mu}$ in (3) is also gauge-invariant, and so its conserved charge corresponds to an physical observable.

Hamiltonian Formalism:

In the pure electromagnetic field section, one finds that the gauge potential $A^{0}$ does not appear in the Lagrangian density. Thus, as a second order formalism its Lagrangian is singular. The canonical momentum of $A^{i}$ is given by $$\Pi^{i}=\frac{\partial\mathcal{L}}{\partial\dot{A}_{i}}=-\dot{A}^{i}+\partial^{i}A^{0}=E^{i}.$$

At the moment, to make the Lorentz invariance manifest, one can define the naive "Poisson Bracket" on the enlarged (unphysical) phase space as $$\left\{F(t),G(t)\right\}_{PB}=\int ds\int d^{3}\mathbf{x}\left\{\frac{\delta F(t)}{\delta A^{\mu}(s,\mathbf{x})}\frac{\delta G(t)}{\delta\Pi_{\mu}(s,\mathbf{x})}-\frac{\delta F(t)}{\delta\Pi_{\mu}(s,\mathbf{x})}\frac{\delta G(t)}{\delta A^{\mu}(s,\mathbf{x})}\right\}, \tag{4.a}$$

where one should keep in mind that $A^{0}(\mathbf{x})$ and $\Pi_{0}(\mathbf{x})$ are unphysical degrees of freedom, which should be set to zero in the end.

Then, the Hamiltonian for the pure electromagnetic field is \begin{align} H_{\mathrm{em}}&=\int d^{3}\mathbf{x}\left(\Pi^{i}\dot{A}_{i}-\mathcal{L}_{\mathrm{em}}\right) \\ &=\int d^{3}\mathbf{x}\left(\Pi_{i}(\partial^{0}A^{i}-\partial^{i}A^{0})+\Pi_{i}\partial^{i}A^{0}-\mathcal{L}_{\mathrm{em}}\right) \\ &=\int d^{3}\mathbf{x}\left(-\Pi_{i}E^{i}+\partial_{i}(\Pi^{i}A^{0})-A^{0}\partial_{i}\Pi^{i}-\mathcal{L}_{\mathrm{em}}\right) \\ &=\int d^{3}\mathbf{x}\left(\frac{1}{2}|\mathbf{\Pi}|^{2}+\frac{1}{2}|\nabla\times\mathbf{A}|^{2}-A^{0}\nabla\cdot\mathbf{\Pi}\right). \end{align}

In the scalar sector, the canonical momenta are $$\Pi=\frac{\partial\mathcal{L}}{\partial\dot{\phi}}=(D_{0}\phi)^{\ast}\quad\mathrm{and}\quad\Pi^{\ast}=\frac{\partial\mathcal{L}}{\partial\dot{\phi}^{\ast}}=D_{0}\phi.$$

The Poisson bracket is defined as \begin{equation} \left\{F(t),G(t)\right\}_{PB} \\ =\int ds\int d^{3}\mathbf{x}\left\{\frac{\delta F(t)}{\delta\phi(s,\mathbf{x})}\frac{\delta G(t)}{\delta\Pi(s,\mathbf{x})}-\frac{\delta F(t)}{\delta\Pi(s,\mathbf{x})}\frac{\delta G(t)}{\delta\phi(s,\mathbf{x})}\right\} \\ +\int ds\int d^{3}\mathbf{x}\left\{\frac{\delta F(t)}{\delta\phi^{\ast}(s,\mathbf{x})}\frac{\delta G(t)}{\delta\Pi^{\ast}(s,\mathbf{x})}-\frac{\delta F(t)}{\delta\Pi^{\ast}(s,\mathbf{x})}\frac{\delta G(t)}{\delta\phi^{\ast}(s,\mathbf{x})}\right\}. \tag{4.b} \end{equation}

Then, the scalar Hamiltonian is given by \begin{align} H_{\mathrm{scalar}}&=\int d^{3}\mathbf{x}\left(\Pi\dot{\phi}+\dot{\phi}^{\ast}\Pi^{\ast}-\mathcal{L}\right) \\ &=\int d^{3}\mathbf{x}\left((D_{0}\phi)^{\ast}\dot{\phi}+\dot{\phi}^{\ast}(D_{0}\phi)-(D_{0}\phi)^{\ast}(D_{0}\phi)+(\mathbf{D}\phi)^{\ast}\cdot(\mathbf{D}\phi)+m^{2}|\phi|^{2}\right) \\ &=\int d^{3}\mathbf{x}\left(\Pi^{\ast}\Pi+A_{0}\rho+(\mathbf{D}\phi)^{\ast}\cdot(\mathbf{D}\phi)+m^{2}|\phi|^{2}\right), \end{align}

where $\rho=J^{0}=i(\Pi\phi-\phi^{\ast}\Pi^{\ast})$, and $(\mathbf{D}\phi)_{k}=D_{k}\phi$.

On the other hand, one has \begin{align} &\,\,\,\,\,\,\,(\mathbf{D}\phi)^{\ast}\cdot(\mathbf{D}\phi)=(\nabla\phi^{\ast}+i\mathbf{A}\phi^{\ast})\cdot(\mathbf{D}\phi)=(\mathbf{D}\phi)^{\ast}\cdot(\nabla\phi-i\mathbf{A}\phi) \\ &=\frac{1}{2}\left[(\nabla\phi^{\ast}+i\mathbf{A}\phi^{\ast})\cdot(\mathbf{D}\phi)+(\mathbf{D}\phi)^{\ast}\cdot(\nabla\phi-i\mathbf{A}\phi)\right] \\ &=\frac{1}{2}\left[(\nabla\phi^{\ast})\cdot(\mathbf{D}\phi)+(\mathbf{D}\phi)^{\ast}\cdot(\nabla\phi)+\mathbf{A}\cdot\mathbf{J}\right] \\ &=(\nabla\phi^{\ast})\cdot(\nabla\phi)+\frac{i}{2}\mathbf{A}\cdot\left[\phi^{\ast}(\nabla\phi)-(\nabla\phi^{\ast})\phi\right]+\frac{1}{2}\mathbf{A}\cdot\mathbf{J} \\ &=(\nabla\phi^{\ast})\cdot(\nabla\phi)+\mathbf{A}\cdot\mathbf{J}-|\mathbf{A}|^{2}|\phi|^{2}, \end{align}

where $\mathbf{J}=i\left[\phi^{\ast}(\mathbf{D}\phi)-(\mathbf{D}\phi)^{\ast}\phi\right]$.

Therefore, the Hamiltonian in the scalar sector is $$H_{\mathrm{scalar}}=\int d^{3}\mathbf{x}\left(\Pi^{\ast}\Pi+(m^{2}-|\mathbf{A}|^{2})|\phi|^{2}+(\nabla\phi^{\ast})\cdot(\nabla\phi)+A_{0}\rho+\mathbf{A}\cdot\mathbf{J}\right),$$

whose gauge invariance can be easily verified.

So the total Hamiltonian of the scalar QED is given by $$H=H_{\mathrm{em}}+H_{\mathrm{scalar}} \tag{5}$$ $$=\int d^{3}\mathbf{x}\left(\frac{1}{2}|\mathbf{\Pi}|^{2}+\frac{1}{2}|\nabla\times\mathbf{A}|^{2}-A^{0}(\nabla\cdot\mathbf{\Pi}-\rho)+|\Pi|^{2}+(m^{2}-|\mathbf{A}|^{2})|\phi|^{2}+(\nabla\phi^{\ast})\cdot(\nabla\phi)+\mathbf{A}\cdot\mathbf{J}\right),$$

which is gauge invariant. Now we can introduce the unphysical degree of freedom $\Pi^{0}$ because it is set to be zero anyway. Then, one can write (5) as \begin{align} H&=\int d^{3}\mathbf{x}\left(\frac{1}{2}\Pi_{\mu}\Pi^{\mu}+\frac{1}{2}|\nabla\times\mathbf{A}|^{2}-A^{0}(\partial_{\mu}\Pi^{\mu}-\rho)\right) \\ &+\int d^{3}\mathbf{x}\left(|\Pi|^{2}+(m^{2}-|\mathbf{A}|^{2})|\phi|^{2}+(\nabla\phi^{\ast})\cdot(\nabla\phi)+\mathbf{A}\cdot\mathbf{J}\right). \end{align}

Claim: The $U(1)$-gauge invariance is generated by the following functional: $$\mathcal{G}[\epsilon](t)=\int d^{3}\mathbf{x}\left(\partial_{\mu}\Pi^{\mu}(t,\mathbf{x})-\rho(t,\mathbf{x})\right)\epsilon(t,\mathbf{x}).$$

Indeed, using the Poisson bracket (4), one obtains the fundamental canonical relation $$\left\{A_{\mu}(\mathbf{x}),\Pi_{\nu}(\mathbf{y})\right\}_{PB}=-g_{\mu\nu}\delta(\mathbf{x}-\mathbf{y})\quad\mathrm{and}\quad\left\{\phi(\mathbf{x}),\Pi(\mathbf{y})\right\}_{PB}=-\delta(\mathbf{x}-\mathbf{y}). \tag{6}$$

Notice that the above Poisson bracket is in fact ill-defined since it's incompatible with the Gauss constraint $\partial_{\mu}\Pi^{\mu}(\mathbf{x})-\rho(\mathbf{x})=0$ from variation with respect to $A^{0}$, and since $\Pi_{0}=0$, it is impossible to satisfy. But this is precisely because we started from the "enlarged" phase space that contains unphysical degrees of freedom due to the gauge redundancies. In fact, the constraint equations must not be substituted into the Poisson brackets. They can only be imposed after computing the Poisson brackets. Then, these constraints projects the result onto the reduced (physical) phase space.

Using relation (6), it's easy to see that \begin{align} &\delta_{\mathcal{G}}A_{\mu}(\mathbf{y})=\left\{A_{\mu}(\mathbf{y}),\mathcal{G}[\epsilon]\right\}_{PB}=\partial_{\mu}\epsilon(\mathbf{y}) \\ &\delta_{\mathcal{G}}\Pi_{\mu}(\mathbf{y})=\left\{\Pi_{\mu}(\mathbf{y}),\mathcal{G}[\epsilon]\right\}_{PB}=0 \\ &\delta_{\mathcal{G}}\phi(\mathbf{y})=\left\{\phi(\mathbf{y}),\mathcal{G}[\epsilon]\right\}_{PB}=i\epsilon(\mathbf{y})\phi(\mathbf{y}) \\ &\delta_{\mathcal{G}}\Pi(\mathbf{y})=\left\{\Pi(\mathbf{y}),\mathcal{G}[\epsilon]\right\}_{PB}=i\epsilon(\mathbf{y})\Pi(\mathbf{y}) \end{align}

which are indeed infinitesimal gauge transformations in the "enlarged" phase space.

So the answer to your question is Yes. The generator $\mathcal{G}[\epsilon]$ simultaneously generates gauge transformations on both gauge fields and the scalar fields.

TO BE CONTINUED